800 18. Radiation from Apertures - Rutgers ECEorfanidi/ewa/ch18.pdf · 18 Radiation from Apertures 18.1 Field Equivalence Principle The radiation ﬁelds from aperture antennas, such

18Radiation from Apertures

18.1 Field Equivalence Principle

The radiation fields from aperture antennas, such as slots, open-ended waveguides,horns, reflector and lens antennas, are determined from the knowledge of the fieldsover the aperture of the antenna.

The aperture fields become the sources of the radiated fields at large distances. Thisis a variation of the Huygens-Fresnel principle, which states that the points on eachwavefront become the sources of secondary spherical waves propagating outwards andwhose superposition generates the next wavefront.

Let Ea,Ha be the tangential fields over an aperture A, as shown in Fig. 18.1.1. Thesefields are assumed to be known and are produced by the sources to the left of the screen.The problem is to determine the radiated fields E(r),H(r) at some far observation point.

The radiated fields can be computed with the help of the field equivalence principle[1288–1295,1685], which states that the aperture fields may be replaced by equivalentelectric and magnetic surface currents, whose radiated fields can then be calculated usingthe techniques of Sec. 15.10. The equivalent surface currents are:

J s = n×Ha

Jms = −n× Ea

(electric surface current)

(magnetic surface current)(18.1.1)

where n is a unit vector normal to the surface and on the side of the radiated fields.Thus, it becomes necessary to consider Maxwell’s equations in the presence of mag-

netic currents and derive the radiation fields from such currents.The screen in Fig. 18.1.1 is an arbitrary infinite surface over which the tangential

fields are assumed to be zero. This assumption is not necessarily consistent with theradiated field solutions, that is, Eqs. (18.4.9). A consistent calculation of the fields tothe right of the aperture plane requires knowledge of the fields over the entire apertureplane (screen plus aperture.)

However, for large apertures (with typical dimension much greater than a wave-length), the approximation of using the fields Ea,Ha only over the aperture to calculatethe radiation patterns is fairly adequate, especially in predicting the main-lobe behaviorof the patterns.

800 18. Radiation from Apertures

Fig. 18.1.1 Radiated fields from an aperture.

The screen can also be a perfectly conducting surface, such as a ground plane, onwhich the aperture opening has been cut. In reflector antennas, the aperture itself isnot an opening, but rather a reflecting surface. Fig. 18.1.2 depicts some examples ofscreens and apertures: (a) an open-ended waveguide over an infinite ground plane, (b)an open-ended waveguide radiating into free space, and (c) a reflector antenna.

Fig. 18.1.2 Examples of aperture planes.

There are two alternative forms of the field equivalence principle, which may be usedwhen only one of the aperture fields Ea or Ha is available. They are:

J s = 0

Jms = −2(n× Ea)(perfect electric conductor) (18.1.2)

J s = 2(n×Ha)

Jms = 0(perfect magnetic conductor) (18.1.3)

They are appropriate when the screen is a perfect electric conductor (PEC) on whichEa = 0, or when it is a perfect magnetic conductor (PMC) on which Ha = 0. On theaperture, both Ea and Ha are non-zero.

18.2. Magnetic Currents and Duality 801

Using image theory, the perfect electric (magnetic) conducting screen can be elimi-nated and replaced by an image magnetic (electric) surface current, doubling its valueover the aperture. The image field causes the total tangential electric (magnetic) field tovanish over the screen.

If the tangential fields Ea,Ha were known over the entire aperture plane (screen plusaperture), the three versions of the equivalence principle would generate the same radi-ated fields. But because we consider Ea,Ha only over the aperture, the three versionsgive slightly different results.

In the case of a perfectly conducting screen, the calculated radiation fields (18.4.10)using the equivalent currents (18.1.2) are consistent with the boundary conditions onthe screen.

We provide a justification of the field equivalence principle (18.1.1) in Sec. 18.10using vector diffraction theory and the Stratton-Chu and Kottler formulas. The modifiedforms (18.1.2) and (18.1.3) are justified in Sec. 19.2 where we derive them in two ways:one, using the plane-wave-spectrum representation, and two, using the Franz formulasin conjunction with the extinction theorem discussed in Sec. 18.11, and discuss alsotheir relationship to Rayleigh-Sommerfeld diffraction theory of Sec. 19.1.

18.2 Magnetic Currents and Duality

Next, we consider the solution of Maxwell’s equations driven by the ordinary electriccharge and current densities ρ, J, and in addition, by the magnetic charge and currentdensities ρm, Jm.

Although ρm, Jm are fictitious, the solution of this problem will allow us to identifythe equivalent magnetic currents to be used in aperture problems, and thus, establishthe field equivalence principle. The generalized form of Maxwell’s equations is:

∇∇∇×H = J+ jωεE

∇∇∇ · E = 1

ερ

∇∇∇× E = −Jm − jωμH

∇∇∇ ·H = 1

μρm

(18.2.1)

There is now complete symmetry, or duality, between the electric and the magneticquantities. In fact, it can be verified easily that the following duality transformationleaves the set of four equations invariant :

E −→ HH −→ −Eε −→ μμ −→ ε

J −→ Jmρ −→ ρm

Jm −→ −Jρm −→ −ρ

A −→ Amϕ −→ ϕm

Am −→ −Aϕm −→ −ϕ

(duality) (18.2.2)

where ϕ,A and ϕm,Am are the corresponding scalar and vector potentials introducedbelow. These transformations can be recognized as a special case (for α = π/2) of thefollowing duality rotations, which also leave Maxwell’s equations invariant:


[E ′ ηJ ′ ηρ′

ηH ′ J ′m ρ′m

]=[

cosα sinα− sinα cosα

][E ηJ ηρ

ηH Jm ρm

](18.2.3)

Under the duality transformations (18.2.2), the first two of Eqs. (18.2.1) transforminto the last two, and conversely, the last two transform into the first two.

A useful consequence of duality is that if one has obtained expressions for the elec-tric field E, then by applying a duality transformation one can generate expressions forthe magnetic field H. We will see examples of this property shortly.

The solution of Eq. (18.2.1) is obtained in terms of the usual scalar and vector po-tentials ϕ,A, as well as two new potentials ϕm,Am of the magnetic type:

E = −∇∇∇ϕ− jωA− 1

ε∇∇∇× Am

H = −∇∇∇ϕm − jωAm + 1

μ∇∇∇× A

(18.2.4)

The expression for H can be derived from that of E by a duality transformation ofthe form (18.2.2). The scalar and vector potentials satisfy the Lorenz conditions andHelmholtz wave equations:

∇∇∇ · A+ jωεμϕ = 0

∇2ϕ+ k2ϕ = −ρε

∇2A+ k2A = −μ J

and

∇∇∇ · Am + jωεμϕm = 0

∇2ϕm + k2ϕm = −ρmμ∇2Am + k2Am = −ε Jm

(18.2.5)

The solutions of the Helmholtz equations are given in terms of G(r− r′)= e−jk|r−r′|

4π|r− r′| :

ϕ(r) =∫V

1

ερ(r′)G(r− r′)dV′,

A(r) =∫Vμ J(r′)G(r− r′)dV′,

ϕm(r) =∫V

1

μρm(r′)G(r− r′)dV′

Am(r) =∫Vε Jm(r′)G(r− r′)dV′

(18.2.6)

where V is the volume over which the charge and current densities are nonzero. Theobservation point r is taken to be outside this volume. Using the Lorenz conditions, thescalar potentials may be eliminated in favor of the vector potentials, resulting in thealternative expressions for Eq. (18.2.4):

E = 1

jωμε[∇∇∇(∇∇∇ · A)+k2A

]− 1

ε∇∇∇× Am

H = 1

jωμε[∇∇∇(∇∇∇ · Am)+k2Am

]+ 1

μ∇∇∇× A

(18.2.7)

These may also be written in the form of Eq. (15.3.9):

E = 1

jωμε[∇∇∇× (∇∇∇× A)−μ J]−1

ε∇∇∇× Am

H = 1

jωμε[∇∇∇× (∇∇∇× Am)−ε Jm]+ 1

μ∇∇∇× A

(18.2.8)

18.3. Radiation Fields from Magnetic Currents 803

Replacing A,Am in terms of Eq. (18.2.6), we may express the solutions (18.2.7) di-rectly in terms of the current densities:

E = 1

jωε

∫V

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′

H = 1

jωμ

∫V

[k2JmG+ (Jm ·∇∇∇′)∇∇∇′G+ jωμ J×∇∇∇′G]dV′

(18.2.9)

Alternatively, if we also use the charge densities, we obtain from (18.2.4):

E =∫V

[−jωμ JG+ ρε∇∇∇′G− Jm ×∇∇∇′G

]dV′

H =∫V

[−jωε JmG+ ρmμ ∇∇∇′G+ J×∇∇∇′G]dV′(18.2.10)

18.3 Radiation Fields from Magnetic Currents

The radiation fields of the solutions (18.2.7) can be obtained by making the far-fieldapproximation, which consists of the replacements:

G(r− r′)= e−jk|r−r′|

4π|r− r′| �e−jkr

4πrejk·r

′and ∇∇∇ � −jk (18.3.1)

where k = kr. Then, the vector potentials of Eq. (18.2.6) take the simplified form:

A(r)= μ e−jkr

4πrF(θ,φ) , Am(r)= ε e

−jkr

4πrFm(θ,φ) (18.3.2)

where the radiation vectors are the Fourier transforms of the current densities:

F(θ,φ) =∫V

J(r′)ejk·r′dV′

Fm(θ,φ) =∫V

Jm(r′)ejk·r′dV′

(radiation vectors) (18.3.3)

Setting J = Jm = 0 in Eq. (18.2.8) because we are evaluating the fields far from thecurrent sources, and using the approximation ∇∇∇ = −jk = −jkr, and the relationshipk/ε =ωη, we find the radiated E and H fields:

E = −jω[r× (A× r)−η r× Am] = −jk e−jkr

4πrr× [ηF× r− Fm

]

H = − jωη[η r× (Am × r)+r× A

] = − jkηe−jkr

4πrr× [ηF+ Fm × r

] (18.3.4)

These generalize Eq. (15.10.2) to magnetic currents. As in Eq. (15.10.3), we have:

H = 1

ηr× E (18.3.5)


Noting that r× (F× r)= θθθFθ + φφφFφ and r× F = φφφFθ − θθθFφ, and similarly for Fm,we find for the polar components of Eq. (18.3.4):

E = −jk e−jkr

4πr[θθθ(ηFθ + Fmφ)+φφφ(ηFφ − Fmθ)

]

H = − jkηe−jkr

4πr[−θθθ(ηFφ − Fmθ)+φφφ(ηFθ + Fmφ)]

(18.3.6)

The Poynting vector is given by the generalization of Eq. (16.1.1):

PPP = 1

2Re(E×H∗)= r

k2

32π2ηr2

[|ηFθ + Fmφ|2 + |ηFφ − Fmθ|2] = rPr (18.3.7)

and the radiation intensity:

U(θ,φ)= dPdΩ

= r2Pr = k2

32π2η[|ηFθ + Fmφ|2 + |ηFφ − Fmθ|2] (18.3.8)

18.4 Radiation Fields from Apertures

For an aperture antenna with effective surface currents given by Eq. (18.1.1), the volumeintegrations in Eq. (18.2.9) reduce to surface integrations over the aperture A:

E = 1

jωε

∫A

[(J s ·∇∇∇′)∇∇∇′G+ k2J s G− jωε Jms ×∇∇∇′G

]dS′

H = 1

jωμ

∫A

[(Jms ·∇∇∇′)∇∇∇′G+ k2Jms G+ jωμ J s ×∇∇∇′G

]dS′

(18.4.1)

and, explicitly in terms of the aperture fields shown in Fig. 18.1.1:

E = 1

jωε

∫A

[(n×Ha)·∇∇∇′(∇∇∇′G)+k2(n×Ha)G+ jωε(n× Ea)×∇∇∇′G

]dS′

H = 1

jωμ

∫A

[−(n× Ea)·∇∇∇′(∇∇∇′G)−k2(n× Ea)G+ jωμ(n×Ha)×∇∇∇′G]dS′

(18.4.2)These are known as Kottler’s formulas [1293–1298,1287,1299–1302,1324]. We derive

them in Sec. 18.12. The equation for H can also be obtained from that of E by theapplication of a duality transformation, that is, Ea → Ha, Ha → −Ea and ε→ μ, μ→ ε.

In the far-field limit, the radiation fields are still given by Eq. (18.3.6), but now theradiation vectors are given by the two-dimensional Fourier transform-like integrals overthe aperture:

F(θ,φ) =∫A

J s(r′)ejk·r′dS′ =

∫A

n×Ha(r′)ejk·r′dS′

Fm(θ,φ) =∫A

Jms(r′)ejk·r′dS′ = −

∫A

n× Ea(r′)ejk·r′dS′

(18.4.3)

18.4. Radiation Fields from Apertures 805

Fig. 18.4.1 Radiation fields from an aperture.

Fig. 18.4.1 shows the polar angle conventions, where we took the origin to be some-where in the middle of the aperture A.

The aperture surface A and the screen in Fig. 18.1.1 can be arbitrarily curved. How-ever, a common case is to assume that they are both flat. Then, Eqs. (18.4.3) becomeordinary 2-d Fourier transform integrals. Taking the aperture plane to be the xy-planeas in Fig. 18.1.1, the aperture normal becomes n = z, and thus, it can be taken out ofthe integrands. Setting dS′ = dx′dy′, we rewrite Eq. (18.4.3) in the form:

F(θ,φ) =∫A

J s(r′)ejk·r′dx′dy′ = z×

∫A

Ha(r′)ejk·r′dx′dy′

Fm(θ,φ) =∫A

Jms(r′)ejk·r′dx′dy′ = −z×

∫A

Ea(r′)ejk·r′dx′dy′

(18.4.4)

where ejk·r′ = ejkxx′+jkyy′ and kx = k cosφ sinθ, ky = k sinφ sinθ. It proves conve-nient then to introduce the two-dimensional Fourier transforms of the aperture fields:

f(θ,φ)=∫A

Ea(r′)ejk·r′dx′dy′ =

∫A

Ea(x′, y′)ejkxx′+jkyy′ dx′dy′

g(θ,φ)=∫A

Ha(r′)ejk·r′dx′dy′ =

∫A

Ha(x′, y′)ejkxx′+jkyy′ dx′dy′

(18.4.5)

Then, the radiation vectors become:

F(θ,φ) = z× g(θ,φ)

Fm(θ,φ) = −z× f(θ,φ)(18.4.6)

Because Ea,Ha are tangential to the aperture plane, they can be resolved into theircartesian components, for example, Ea = xEax + yEay. Then, the quantities f,g can beresolved in the same way, for example, f = x fx + y fy. Thus, we have:


F = z× g = z× (xgx + ygy)= ygx − xgy

Fm = −z× f = −z× (x fx + y fy)= x fy − y fx(18.4.7)

The polar components of the radiation vectors are determined as follows:

Fθ = θθθ · F = θθθ · (ygx − xgy)= gx sinφ cosθ− gy cosφ cosθ

where we read off the dot products (θθθ · x) and (θθθ · y) from Eq. (15.8.3). The remainingpolar components are found similarly, and we summarize them below:

Fθ = − cosθ(gy cosφ− gx sinφ)

Fφ = gx cosφ+ gy sinφ

Fmθ = cosθ(fy cosφ− fx sinφ)

Fmφ = −(fx cosφ+ fy sinφ)

(18.4.8)

It follows from Eq. (18.3.6) that the radiated E-field will be:

Eθ = jk e−jkr

4πr[(fx cosφ+ fy sinφ)+η cosθ(gy cosφ− gx sinφ)

]

Eφ = jk e−jkr

4πr[cosθ(fy cosφ− fx sinφ)−η(gx cosφ+ gy sinφ)

] (18.4.9)

The radiation fields resulting from the alternative forms of the field equivalenceprinciple, Eqs. (18.1.2) and (18.1.3), are obtained from Eq. (18.4.9) by removing the g- orthe f -terms and doubling the remaining term. We have for the PEC case:

Eθ = 2jke−jkr

4πr[fx cosφ+ fy sinφ

]

Eφ = 2jke−jkr

4πr[cosθ(fy cosφ− fx sinφ)

] (18.4.10)

and for the PMC case:

Eθ = 2jke−jkr

4πr[η cosθ(gy cosφ− gx sinφ)

]

Eφ = 2jke−jkr

4πr[−η(gx cosφ+ gy sinφ)

] (18.4.11)

In all three cases, the radiated magnetic fields are obtained from:

Hθ = − 1

ηEφ , Hφ = 1

ηEθ (18.4.12)

We note that Eq. (18.4.9) is the average of Eqs. (18.4.10) and (18.4.11). Also, Eq. (18.4.11)is the dual of Eq. (18.4.10). Indeed, using Eq. (18.4.12), we obtain the following H-components for Eq. (18.4.11), which can be derived from Eq. (18.4.10) by the dualitytransformation Ea → Ha or f → g , that is,

18.5. Huygens Source 807

Hθ = 2jke−jkr

4πr[gx cosφ+ gy sinφ

]

Hφ = 2jke−jkr

4πr[cosθ(gy cosφ− gx sinφ)

] (18.4.13)

At θ = 90o, the components Eφ, Hφ become tangential to the aperture screen. Wenote that because of the cosθ factors, Eφ (resp. Hφ) will vanish in the PEC (resp. PMC)case, in accordance with the boundary conditions.

18.5 Huygens Source

The aperture fields Ea,Ha are referred to as Huygens source if at all points on theaperture they are related by the uniform plane-wave relationship:

Ha = 1

ηn× Ea (Huygens source) (18.5.1)

where η is the characteristic impedance of vacuum.For example, this is the case if a uniform plane wave is incident normally on the

aperture plane from the left, as shown in Fig. 18.5.1. The aperture fields are assumed tobe equal to the incident fields, Ea = Einc and Ha = Hinc, and the incident fields satisfyHinc = z× Einc/η.

Fig. 18.5.1 Uniform plane wave incident on an aperture.

The Huygens source condition is not always satisfied. For example, if the uniformplane wave is incident obliquely on the aperture, then η must be replaced by the trans-verse impedance ηT, which depends on the angle of incidence and the polarization ofthe incident wave as discussed in Sec. 7.2.

Similarly, if the aperture is the open end of a waveguide, then ηmust be replaced bythe waveguide’s transverse impedance, such as ηTE or ηTM, depending on the assumedwaveguide mode. On the other hand, if the waveguide ends are flared out into a hornwith a large aperture, then Eq. (18.5.1) is approximately valid.


The Huygens source condition implies the same relationship for the Fourier trans-forms of the aperture fields, that is, (with n = z)

g = 1

ηn× f ⇒ gx = − 1

ηfy , gy = 1

ηfx (18.5.2)

Inserting these into Eq. (18.4.9) we may express the radiated electric field in termsof f only. We find:

Eθ = jk e−jkr

2πr1+ cosθ

2

[fx cosφ+ fy sinφ

]

Eφ = jk e−jkr

2πr1+ cosθ

2

[fy cosφ− fx sinφ

] (18.5.3)

The factor (1+cosθ)/2 is known as an obliquity factor. The PEC case of Eq. (18.4.10)remains unchanged for a Huygens source, but the PMC case becomes:

Eθ = jk e−jkr

2πrcosθ

[fx cosφ+ fy sinφ

]

Eφ = jk e−jkr

2πr[fy cosφ− fx sinφ

] (18.5.4)

We may summarize all three cases by the single formula:

Eθ = jk e−jkr

2πrcθ[fx cosφ+ fy sinφ

]

Eφ = jk e−jkr

2πrcφ[fy cosφ− fx sinφ

] (fields from Huygens source) (18.5.5)

where the obliquity factors are defined in the three cases:[cθcφ

]= 1

2

[1+ cosθ1+ cosθ

],[

1cosθ

],[

cosθ1

](obliquity factors) (18.5.6)

We note that the first is the average of the last two. The obliquity factors are equal tounity in the forward direction θ = 0o and vary little for near-forward angles. Therefore,the radiation patterns predicted by the three methods are very similar in their mainlobebehavior.

In the case of a modified Huygens source that replaces η by ηT, Eqs. (18.5.5) retaintheir form. The aperture fields and their Fourier transforms are now assumed to berelated by:

Ha = 1

ηTz× Ea ⇒ g = 1

ηTz× f (18.5.7)

Inserting these into Eq. (18.4.9), we obtain the modified obliquity factors :

cθ = 1

2[1+K cosθ] , cφ = 1

2[K + cosθ] , K = η

ηT(18.5.8)

18.6. Directivity and Effective Area of Apertures 809

18.6 Directivity and Effective Area of Apertures

For any aperture, given the radiation fields Eθ, Eφ of Eqs. (18.4.9)–(18.4.11), the corre-sponding radiation intensity is:

U(θ,φ)= dPdΩ

= r2Pr = r2 1

2η[|Eθ|2 + |Eφ|2] = r2 1

2η|E(θ,φ)|2 (18.6.1)

Because the aperture radiates only into the right half-space 0 ≤ θ ≤ π/2, the totalradiated power and the effective isotropic radiation intensity will be:

Prad =∫ π/2

0

∫ 2π

0U(θ,φ)dΩ , UI = Prad

4π(18.6.2)

The directive gain is computed by D(θ,φ)= U(θ,φ)/UI, and the normalized gainby g(θ,φ)= U(θ,φ)/Umax. For a typical aperture, the maximum intensity Umax istowards the forward direction θ = 0o. In the case of a Huygens source, we have:

U(θ,φ)= k2

8π2η[c2θ|fx cosφ+ fy sinφ|2 + c2

φ|fy cosφ− fx sinφ|2] (18.6.3)

Assuming that the maximum is towards θ = 0o, then cθ = cφ = 1, and we find forthe maximum intensity:

Umax = k2

8π2η[|fx cosφ+ fy sinφ|2 + |fy cosφ− fx sinφ|2]θ=0

= k2

8π2η[|fx|2 + |fy|2]θ=0 =

k2

8π2η|f |2max

where |f|2max =[|fx|2 + |fy|2]θ=0. Setting k = 2π/λ, we have:

Umax = 1

2λ2η|f |2max (18.6.4)

It follows that the normalized gain will be:

g(θ,φ)= c2θ|fx cosφ+ fy sinφ|2 + c2

φ|fy cosφ− fx sinφ|2|f |2max

(18.6.5)

In the case of Eq. (18.4.9) with cθ = cφ = (1+ cosθ)/2, this simplifies further into:

g(θ,φ)= c2θ|fx|2 + |fy|2|f |2max

=(

1+ cosθ2

)2 |f(θ,φ)|2|f |2max

(18.6.6)

The square root of the gain is the (normalized) field strength:

|E(θ,φ)||E |max

=√g(θ,φ) =

(1+ cosθ

2

) |f(θ,φ)||f |max

(18.6.7)

The power computed by Eq. (18.6.2) is the total power that is radiated outwards froma half-sphere of large radius r. An alternative way to compute Prad is to invoke energy


conservation and compute the total power that flows into the right half-space throughthe aperture. Assuming a Huygens source, we have:

Prad =∫APz dS′ = 1

2

∫A

z · Re[Ea ×H∗a

]dS′ = 1

2η

∫A|Ea(r′)|2dS′ (18.6.8)

Because θ = 0 corresponds to kx = ky = 0, it follows from the Fourier transformdefinition (18.4.5) that:

|f|2max =∣∣∣∣∫A

Ea(r′)ejk·r′dS′

∣∣∣∣2

kx=ky=0=∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

Therefore, the maximum intensity is given by:

Umax = 1

2λ2η|f |2max =

1

2λ2η

∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

(18.6.9)

Dividing (18.6.9) by (18.6.8), we find the directivity:

Dmax = 4πUmax

Prad= 4πλ2

∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

∫A|Ea(r′)|2dS′

= 4πAeff

λ2(directivity) (18.6.10)

It follows that the maximum effective area of the aperture is:

Aeff =

∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

∫A|Ea(r′)|2dS′

≤ A (effective area) (18.6.11)

and the aperture efficiency :

ea = Aeff

A=

∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

A∫A|Ea(r′)|2dS′

≤ 1 (aperture efficiency) (18.6.12)

The inequalities in Eqs. (18.6.11) and (18.6.12) can be thought of as special cases ofthe Cauchy-Schwarz inequality. It follows that equality is reached whenever Ea(r′) isuniform over the aperture, that is, independent of r′.

Thus, uniform apertures achieve the highest directivity and have effective areas equalto their geometrical areas.

Because the integrand in the numerator of ea depends both on the magnitude and thephase of Ea, it proves convenient to separate out these effects by defining the aperturetaper efficiency or loss, eatl, and the phase error efficiency or loss, epel, as follows:

eatl =

∣∣∣∣∫A|Ea(r′)|dS′

∣∣∣∣2

A∫A|Ea(r′)|2dS′

, epel =

∣∣∣∣∫A

Ea(r′)dS′∣∣∣∣2

∣∣∣∣∫A|Ea(r′)|dS′

∣∣∣∣2 (18.6.13)

18.7. Uniform Apertures 811

so that ea becomes the product:

ea = eatl epel (18.6.14)

We note that Eq. (18.6.10) was derived under the assumption that the aperture fieldswere Huygens sources, that is, they were related by, ηHa = z × Ea. This assumptionis not necessarily always true, but may be justified for electrically large apertures, i.e.,with dimensions that are much larger than the wavelength λ.

The correct expression for the directivity is given by Eq. (19.7.10) of Sec. 19.7 andcompared with (18.6.10). Directivities that are much larger (even infinite) than those ofthe uniform case are theoretically possible, but almost impossible to realize in practice.The issues associated with such superdirective apertures are discussed in Sec. 20.22.

18.7 Uniform Apertures

In uniform apertures, the fields Ea,Ha are assumed to be constant over the aperturearea. Fig. 18.7.1 shows the examples of a rectangular and a circular aperture. For con-venience, we will assume a Huygens source.

Fig. 18.7.1 Uniform rectangular and circular apertures.

The field Ea can have an arbitrary direction, with constant x- and y-components,Ea = xE0x + yE0y. Because Ea is constant, its Fourier transform f(θ,φ) becomes:

f(θ,φ)=∫A

Ea(r′)ejk·r′dS′ = Ea

∫Aejk·r

′dS′ ≡ Af(θ,φ)Ea (18.7.1)

where we introduced the normalized scalar quantity:

f(θ,φ)= 1

A

∫Aejk·r

′dS′ (uniform-aperture pattern) (18.7.2)

The quantity f(θ,φ) depends on the assumed geometry of the aperture and it, alone,determines the radiation pattern. Noting that the quantity |Ea| cancels out from the


ratio in the gain (18.6.7) and that f(0,φ)= (1/A)∫A dS′ = 1, we find for the normalizedgain and field strengths:

|E(θ,φ)||E |max

=√g(θ,φ) =

(1+ cosθ

2

)|f(θ,φ)| (18.7.3)

18.8 Rectangular Apertures

For a rectangular aperture of sides a,b, the area integral (18.7.2) is separable in the x-and y-directions:

f(θ,φ)= 1

ab

∫ a/2−a/2

∫ b/2−b/2

ejkxx′+jkyy′ dx′dy′ = 1

a

∫ a/2−a/2

ejkxx′dx′ · 1

b

∫ b/2−b/2

ejkyy′dy′

where we placed the origin of the r′ integration in the middle of the aperture. The aboveintegrals result in the sinc-function patterns:

f(θ,φ)= sin(kxa/2)kxa/2

sin(kyb/2)kyb/2

= sin(πvx)πvx

sin(πvy)πvy

(18.8.1)

where we defined the quantities vx, vy :

vx = 1

2πkxa = 1

2πka sinθ cosφ = a

λsinθ cosφ

vy = 1

2πkyb = 1

2πkb sinθ sinφ = b

λsinθ sinφ

(18.8.2)

The pattern simplifies along the two principal planes, the xz- and yz-planes, corre-sponding to φ = 0o and φ = 90o. We have:

f(θ,0o) = sin(πvx)πvx

= sin((πa/λ)sinθ

)(πa/λ)sinθ

f(θ,90o) = sin(πvy)πvy

= sin((πb/λ)sinθ

)(πb/λ)sinθ

(18.8.3)

Fig. 18.8.1 shows the three-dimensional pattern of Eq. (18.7.3) as a function of theindependent variables vx, vy, for aperture dimensions a = 8λ and b = 4λ. The x, yseparability of the pattern is evident. The essential MATLAB code for generating thisfigure was (note MATLAB’s definition of sinc(x)= sin(πx)/(πx)):

a = 8; b = 4;[theta,phi] = meshgrid(0:1:90, 0:9:360);theta = theta*pi/180; phi = phi*pi/180;

vx = a*sin(theta).*cos(phi);vy = b*sin(theta).*sin(phi);

E = abs((1 + cos(theta))/2 .* sinc(vx) .* sinc(vy));

surfl(vx,vy,E);shading interp; colormap(gray(16));

18.8. Rectangular Apertures 813

−8−4

04

8

−8−4

04

80

0.5

1

xv yv

htgnerts dleif

Fig. 18.8.1 Radiation pattern of rectangular aperture (a = 8λ, b = 4λ).

As the polar angles vary over 0 ≤ θ ≤ 90o and 0 ≤ φ ≤ 360o, the quantities vx andvy vary over the limits −a/λ ≤ vx ≤ a/λ and −b/λ ≤ vy ≤ b/λ. In fact, the physicallyrealizable values of vx, vy are those that lie in the ellipse in the vxvy-plane:

v2xa2+ v

2y

b2≤ 1

λ2(visible region) (18.8.4)

The realizable values of vx, vy are referred to as the visible region. The graph inFig. 18.8.1 restricts the values of vx, vy within that region.

The radiation pattern consists of a narrow mainlobe directed towards the forwarddirection θ = 0o and several sidelobes.

We note the three characteristic properties of the sinc-function patterns: (a) the 3-dB width in v-space is Δvx = 0.886 (the 3-dB wavenumber is vx = 0.443); (b) the firstsidelobe is down by about 13.26 dB from the mainlobe and occurs at vx = 1.4303; and(c) the first null occurs at vx = 1. See Sec. 22.7 for the proof of these results.

The 3-dB width in angle space can be obtained by linearizing the relationship vx =(a/λ)sinθ about θ = 0o, that is, Δvx = (a/λ)Δθ cosθ

∣∣θ=0 = aΔθ/λ. Thus, Δθ =

λΔvx/a. This ignores also the effect of the obliquity factor. It follows that the 3-dBwidths in the two principal planes are (in radians and in degrees):

Δθx = 0.886λa= 50.76o λ

a, Δθy = 0.886

λb= 50.76o λ

b(18.8.5)

The 3-dB angles are θx = Δθx/2 = 25.4o λ/a and θy = Δθy/2 = 25.4o λ/b.Fig. 18.8.2 shows the two principal radiation patterns of Eq. (18.7.3) as functions ofθ, for the case a = 8λ, b = 4λ. The obliquity factor was included, but it makes essen-tially no difference near the mainlobe and first sidelobe region, ultimately suppressingthe response at θ = 90o by a factor of 0.5.

The 3-dB widths are shown on the graphs. The first sidelobes occur at the anglesθa = asin(1.4303λ/a)= 10.30o and θb = asin(1.4303λ/b)= 20.95o.


0 10 20 30 40 50 60 70 80 900

0.5

1

fie

ld s

tren

gth

θ (degrees)

Radiation Pattern for φ = 0o

3 dB

13.26 dB

0 10 20 30 40 50 60 70 80 900

0.5

1

fie

ld s

tren

gth

θ (degrees)

Radiation Pattern for φ φ = 90o

3 dB

13.26 dB

Fig. 18.8.2 Radiation patterns along the two principal planes (a = 8λ, b = 4λ).

For aperture antennas, the gain is approximately equal to the directivity because thelosses tend to be very small. The gain of the uniform rectangular aperture is, therefore,G � D = 4π(ab)/λ2. Multiplying G by Eqs. (18.8.5), we obtain the gain-beamwidthproduct p = GΔθx Δθy = 4π(0.886)2= 9.8646 rad2 = 32 383 deg2. Thus, we have anexample of the general formula (16.3.14) (with the angles in radians and in degrees):

G = 9.8646

Δθx Δθy= 32 383

Δθox Δθo

y(18.8.6)

18.9 Circular Apertures

For a circular aperture of radius a, the pattern integral (18.7.2) can be done convenientlyusing cylindrical coordinates. The cylindrical symmetry implies that f(θ,φ) will beindependent of φ.

Therefore, for the purpose of computing the integral (18.7.2), we may setφ = 0. Wehave then k · r′ = kxx′ = kρ′ sinθ cosφ′. Writing dS′ = ρ′dρ′dφ′, we have:

f(θ)= 1

πa2

∫ a0

∫ 2π

0ejkρ

′ sinθ cosφ′ρ′ dρ′dφ′ (18.9.1)

Theφ′- and ρ′-integrations can be done using the following integral representationsfor the Bessel functions J0(x) and J1(x) [1790]:

J0(x)= 1

2π

∫ 2π

0ejx cosφ′ dφ′ and

∫ 1

0J0(xr)r dr = J1(x)

x(18.9.2)

Then Eq. (18.9.1) gives:

f(θ)= 2J1(ka sinθ)ka sinθ

= 2J1(2πu)

2πu, u = 1

2πka sinθ = a

λsinθ (18.9.3)

This is the well-known Airy pattern [638] for a circular aperture. The function f(θ)is normalized to unity at θ = 0o, because J1(x) behaves like J1(x)� x/2 for small x.

18.9. Circular Apertures 815

Fig. 18.9.1 shows the three-dimensional field pattern (18.7.3) as a function of the in-dependent variables vx = (a/λ)sinθ cosφ and vy = (a/λ)sinθ sinφ, for an apertureradius of a = 3λ. The obliquity factor was not included as it makes little differencenear the main lobe. The MATLAB code for this graph was implemented with the built-infunction besselj:

−3

0

3

−3

0

30

0.5

1

xv yv

htgnerts dleif

Fig. 18.9.1 Radiation pattern of circular aperture (a = 3λ).

a = 3;[theta,phi] = meshgrid(0:1:90, 0:9:360);theta = theta*pi/180; phi = phi*pi/180;

vx = a*sin(theta).*cos(phi);vy = a*sin(theta).*sin(phi);u = a*sin(theta);

E = ones(size(u));i = find(u);E(i) = abs(2*besselj(1,2*pi*u(i))./(2*pi*u(i)));

surfl(vx,vy,E);shading interp; colormap(gray(16));

The visible region is the circle on the vxvy-plane:

v2x + v2

y ≤a2

λ2(18.9.4)

The mainlobe/sidelobe characteristics of f(θ) are as follows. The 3-dB wavenumberis u = 0.2572 and the 3-dB width in u-space is Δu = 2×0.2572 = 0.5144. The first nulloccurs at u = 0.6098 so that the first-null width is Δu = 2×0.6098 = 1.22. The firstsidelobe occurs at u = 0.8174 and its height is |f(u)| = 0.1323 or 17.56 dB below themainlobe. The beamwidths in angle space can be obtained from Δu = a(Δθ)/λ, whichgives for the 3-dB and first-null widths in radians and degrees:

Δθ3dB = 0.5144λa= 29.47o λ

a, Δθnull = 1.22

λa= 70o λ

a(18.9.5)


0 10 20 30 40 50 60 70 80 900

0.5

1

fie

ld s

tren

gth

θ (degrees)

Radiation Pattern of Circular Aperture

3 dB

17.56 dB

Fig. 18.9.2 Radiation pattern of circular aperture (a = 3λ).

The 3-dB angle is θ3dB = Δθ3dB/2 = 0.2572λ/a = 14.74o λ/a and the first-nullangle θnull = 0.6098λ/a. Fig. 18.9.2 shows the radiation pattern of Eq. (18.7.3) as afunction of θ, for the case a = 3λ. The obliquity factor was included.

The graph shows the 3-dB width and the first sidelobe, which occurs at the angleθa =asin(0.817λ/a)= 15.8o. The first null occurs at θnull = asin(0.6098λ/a)= 11.73o,whereas the approximation θnull = 0.6098λ/a gives 11.65o.

The gain-beamwidth product is p = G(Δθ3dB)2= [4π(πa2)/λ2

](0.514λ/a)2=

4π2(0.5144)2= 10.4463 rad2 = 34 293 deg2. Thus, in radians and degrees:

G = 10.4463

(Δθ3dB)2= 34 293

(Δθo3dB)2

(18.9.6)

The first-null angle θnull = 0.6098λ/a is the so-called Rayleigh diffraction limit forthe nominal angular resolution of optical instruments, such as microscopes and tele-scopes. It is usually stated in terms of the diameter D = 2a of the optical aperture:

Δθ = 1.22λD= 70o λ

D(Rayleigh limit) (18.9.7)

18.10 Vector Diffraction Theory

In this section, we provide a justification of the field equivalence principle (18.1.1) andKottler’s formulas (18.4.2) from the point of view of vector diffraction theory. We alsodiscuss the Stratton-Chu and Franz formulas. A historical overview of this subject isgiven in [1302,1324].

In Sec. 18.2, we worked with the vector potentials and derived the fields due toelectric and magnetic currents radiating in an unbounded region. Here, we consider theproblem of finding the fields in a volumeV bounded by a closed surface S and an infinitespherical surface S∞, as shown in Fig. 18.10.1.

The solution of this problem requires that we know the current sources within Vand the electric and magnetic fields tangential to the surface S. The fields E1,H1 and

18.10. Vector Diffraction Theory 817

Fig. 18.10.1 Fields outside a closed surface S.

current sources inside the volume V1 enclosed by S have an effect on the outside onlythrough the tangential fields on the surface.

We start with Maxwell’s equations (18.2.1), which include both electric and magneticcurrents. This will help us identify the effective surface currents and derive the fieldequivalence principle.

Taking the curls of both sides of Ampere’s and Faraday’s laws and using the vectoridentity∇∇∇×(∇∇∇×E)=∇∇∇(∇∇∇·E)−∇2E, we obtain the following inhomogeneous Helmholtzequations (which are duals of each other):

∇2E+ k2E = jωμ J+ 1

ε∇∇∇ρ+∇∇∇× Jm

∇2H+ k2H = jωε Jm + 1

μ∇∇∇ρm −∇∇∇× J

(18.10.1)

We recall that the Green’s function for the Helmholtz equation is:

∇′2G+ k2G = −δ(3)(r− r′) , G(r− r′)= e−jk|r−r′|

4π|r− r′| (18.10.2)

where ∇∇∇′ is the gradient with respect to r′. Applying Green’s second identity given byEq. (C.27) of Appendix C, we obtain:∫

V

[G∇′2E− E∇′2G]dV′ = −

∮S+S∞

[G∂E

∂n′− E

∂G∂n′

]dS′ ,

∂∂n′

= n ·∇∇∇′

whereG and E stand forG(r−r′) and E(r′) and the integration is over r′. The quantity∂/∂n′ is the directional derivative along n. The negative sign in the right-hand sidearises from using a unit vector n that is pointing into the volume V.

The integral over the infinite surface is taken to be zero. This may be justified morerigorously [1295] by assuming that E and H behave like radiation fields with asymptoticform E → const.e−jkr/r and H → r × E/η.† Thus, dropping the S∞ term, and addingand subtracting k2GE in the left-hand side, we obtain:∫

V

[G(∇′2E+ k2E)−E (∇′2G+ k2G)

]dV′ = −

∮S

[G∂E

∂n′− E

∂G∂n′

]dS′ (18.10.3)

†The precise conditions are: r|E| → const. and r|E− ηH× r| → 0 as r →∞.


Using Eq. (18.10.2), the second term on the left may be integrated to give E(r):

−∫V

E(r′) (∇′2G+ k2G)dV′ =∫V

E(r′)δ(3)(r− r′)dV′ = E(r)

where we assumed that r lies in V. This integral is zero if r lies in V1 because then r′

can never be equal to r. For arbitrary r, we may write:

∫V

E(r′)δ(3)(r− r′)dV′ = uV(r)E(r)=⎧⎨⎩E(r), if r ∈ V

0, if r �∈ V (18.10.4)

where uV(r) is the characteristic, or indicator, function of the volume region V:†

uV(r)=⎧⎨⎩1, if r ∈ V

0, if r �∈ V (18.10.5)

We may now solve Eq. (18.10.3) for E(r). In a similar fashion, or, performing a dualitytransformation on the expression for E(r), we also obtain the corresponding magneticfield H(r). Using (18.10.1), we have:

E(r) =∫V

[−jωμG J− 1

εG∇∇∇′ρ−G∇∇∇′ × Jm

]dV′ +

∮S

[E∂G∂n′

−G ∂E

∂n′

]dS′

H(r) =∫V

[−jωεG Jm − 1

μG∇∇∇′ρm +G∇∇∇′ × J

]dV′ +

∮S

[H∂G∂n′

−G ∂H

∂n′

]dS′

(18.10.6)Because of the presence of the particular surface term, we will refer to these as

the Kirchhoff diffraction formulas. Eqs. (18.10.6) can be transformed into the so-calledStratton-Chu formulas [1293–1298,1287,1299–1302,1324]:‡

E(r)=∫V

[−jωμG J+ ρ

ε∇∇∇′G− Jm ×∇∇∇′G

]dV′

+∮S

[−jωμG(n×H)+(n · E)∇∇∇′G+ (n× E)×∇∇∇′G]dS′

H(r)=∫V

[−jωεG Jm + ρmμ ∇∇∇′G+ J×∇∇∇′G

]dV′

+∮S

[jωεG(n× E)+(n ·H)∇∇∇′G+ (n×H)×∇∇∇′G]dS′

(18.10.7)

The proof of the equivalence of (18.10.6) and (18.10.7) is rather involved. Problem18.4 breaks down the proof into its essential steps.

Term by term comparison of the volume and surface integrals in (18.10.7) yields theeffective surface currents of the field equivalence principle:∗

J s = n×H , Jms = −n× E (18.10.8)

†Technically [1301], one must set uV(r)= 1/2, if r lies on the boundary of V, that is, on S.‡See [1289,1296,1302,1324] for earlier work by Larmor, Tedone, Ignatowski, and others.∗Initially derived by Larmor and Love [1302,1324], and later developed fully by Schelkunoff [1288,1290].

18.10. Vector Diffraction Theory 819

Similarly, the effective surface charge densities are:

ρs = ε n · E , ρms = μ n ·H (18.10.9)

Eqs. (18.10.7) may be transformed into the Kottler formulas [1293–1298,1287,1299–1302,1324], which eliminate the charge densities ρ,ρm in favor of the currents J, Jm :

E(r)= 1

jωε

∫V

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′

+ 1

jωε

∮S

[k2G(n×H)+((n×H)·∇∇∇′)∇∇∇′G+ jωε(n× E)×∇∇∇′G]dS′

H(r)= 1

jωμ

∫V

[k2JmG+ (Jm ·∇∇∇′)∇∇∇′G+ jωμ J×∇∇∇′G

]dV′

+ 1

jωμ

∮S

[−k2G(n× E)−((n× E)·∇∇∇′)∇∇∇′G+ jωμ(n×H)×∇∇∇′G]dS′(18.10.10)

The steps of the proof are outlined in Problem 18.5.A related problem is to consider a volume V bounded by the surface S, as shown in

Fig. 18.10.2. The fields inside V are still given by (18.10.7), with n pointing again intothe volume V. If the surface S recedes to infinity, then (18.10.10) reduce to (18.2.9).

Fig. 18.10.2 Fields inside a closed surface S.

Finally, the Kottler formulas may be transformed into the Franz formulas [1298,1287,1299–1301], which are essentially equivalent to Eq. (18.2.8) amended by the vector potentialsdue to the equivalent surface currents:

E(r) = 1

jωμε[∇∇∇× (∇∇∇× (A+ A s)

)− μ J]− 1

ε∇∇∇× (Am + Ams)

H(r) = 1

jωμε[∇∇∇× (∇∇∇× (Am + Ams)

)− ε Jm]+ 1

μ∇∇∇× (A+ A s)

(18.10.11)

where A and Am were defined in Eq. (18.2.6). The new potentials are defined by:

A s(r) =∮Sμ J s(r′)G(r− r′)dS′ =

∮Sμ[n×H(r′)

]G(r− r′)dS′

Ams(r) =∮Sε Jms(r′)G(r− r′)dS′ = −

∮Sε[n× E(r′)

]G(r− r′)dS′

(18.10.12)


Next, we specialize the above formulas to the case where the volume V containsno current sources (J = Jm = 0), so that the E,H fields are given only in terms of thesurface integral terms.

This happens if we choose S in Fig. 18.10.1 such that all the current sources areinside it, or, if in Fig. 18.10.2 we choose S such that all the current sources are outsideit, then, the Kirchhoff, Stratton-Chu, Kottler, and Franz formulas simplify into:

E(r) =∮S

[E∂G∂n′

−G ∂E

∂n′

]dS′

=∮S

[−jωμG(n×H )+(n · E )∇∇∇′G+ (n× E )×∇∇∇′G]dS′

= 1

jωε

∮S

[k2G(n×H )+((n×H )·∇∇∇′)∇∇∇′G+ jωε(n× E )×∇∇∇′G]dS′

= 1

jωε∇∇∇× (∇∇∇×

∮SG(n×H )dS′

)+∇∇∇×∮SG(n× E )dS′

(18.10.13)

H(r) =∮S

[H∂G∂n′

−G ∂H

∂n′

]dS′

=∮S

[jωεG(n× E )+(n ·H )∇∇∇′G+ (n×H )×∇∇∇′G]dS′

= 1

jωμ

∮S

[−k2G(n× E )−((n× E )·∇∇∇′)∇∇∇′G+ jωμ(n×H )×∇∇∇′G]dS′

= − 1

jωμ∇∇∇× (∇∇∇×

∮SG(n× E )dS′

)+∇∇∇×∮SG(n×H )dS′

(18.10.14)where the last equations are the Franz formulas with A = Am = 0.

Fig. 18.10.3 illustrates the geometry of the two cases. Eqs. (18.10.13) and (18.10.14)represent the vectorial formulation of the Huygens-Fresnel principle, according to whichthe tangential fields on the surface can be considered to be the sources of the fields awayfrom the surface.

Fig. 18.10.3 Current sources are outside the field region.

18.11. Extinction Theorem 821

18.11 Extinction Theorem

In all of the equivalent formulas for E(r),H(r), we assumed that r lies within the volumeV. The origin of the left-hand sides in these formulas can be traced to Eq. (18.10.4), andtherefore, if r is not in V but is within the complementary volume V1, then the left-handsides of all the formulas are zero. This does not mean that the fields inside V1 arezero—it only means that the sum of the terms on the right-hand sides are zero.

To clarify these remarks, we consider an imaginary closed surface S dividing allspace in two volumes V1 and V, as shown in Fig. 18.11.1. We assume that there arecurrent sources in both regions V and V1. The surface S1 is the same as S but its unitvector n1 points intoV1, so that n1 = −n. Applying (18.10.10) to the volumeV, we have:

Fig. 18.11.1 Current sources may exist in both V and V1.

1

jωε

∮S


+ 1

jωε

∫V

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′ =

⎧⎨⎩E(r), if r ∈ V

0, if r ∈ V1

The vanishing of the right-hand side when r is in V1 is referred to as an extinctiontheorem.† Applying (18.10.10) to V1, and denoting by E1,H1 the fields in V1, we have:

1

jωε

∮S1

[k2G(n1 ×H1)+

((n1 ×H1)·∇∇∇′

)∇∇∇′G+ jωε(n1 × E1)×∇∇∇′G]dS′

+ 1

jωε

∫V1

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′ =

⎧⎨⎩0, if r ∈ V

E1(r), if r ∈ V1

Because n1 = −n, and on the surface E1 = E and H1 = H, we may rewrite:

− 1

jωε

∮S


+ 1

jωε

∫V1

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′ =

⎧⎨⎩0, if r ∈ V

E1(r), if r ∈ V1

Adding up the two cases and combining the volume integrals into a single one, we obtain:

1

jωε

∫V+V1

[(J ·∇∇∇′)∇∇∇′G+ k2GJ− jωε Jm ×∇∇∇′G

]dV′ =

⎧⎨⎩E(r), if r ∈ V

E1(r), if r ∈ V1

†In fact, it can be used to prove the Ewald-Oseen extinction theorem that we considered in Sec. 15.6.


This is equivalent to Eq. (18.2.9) in which the currents are radiating into unboundedspace. We can also see how the sources within V1 make themselves felt on the outsideonly through the tangential fields at the surface S, that is, for r ∈ V :

1

jωε

∫V1

[k2JG+ (J ·∇∇∇′)∇∇∇′G− jωε Jm ×∇∇∇′G

]dV′

= 1

jωε

∮S


18.12 Vector Diffraction for Apertures

The Kirchhoff diffraction integral, Stratton-Chu, Kottler, and Franz formulas are equiv-alent only for a closed surface S.

If the surface is open, as in the case of an aperture, the four expressions in (18.10.13)and in (18.10.14) are no longer equivalent. In this case, the Kottler and Franz formulasremain equal to each other and give the correct expressions for the fields, in the sensethat the resulting E(r) and H(r) satisfy Maxwell’s equations [1289,1287,1302,1324].

For an open surface S bounded by a contour C, shown in Fig. 18.12.1, the Kottlerand Franz formulas are related to the Stratton-Chu and the Kirchhoff diffraction integralformulas by the addition of some line-integral correction terms [1296]:

E(r)= 1

jωε

∫S


= 1

jωε∇∇∇× (∇∇∇×

∫SG(n×H )dS′

)+∇∇∇×∫SG(n× E )dS′

=∫S

[−jωμG(n×H )+(n · E )∇∇∇′G+ (n× E )×∇∇∇′G]dS′ − 1

jωε

∮C(∇∇∇′G)H · dl

=∫S

[E∂G∂n′

−G ∂E

∂n′

]dS′ −

∮CGE× dl− 1

jωε

∮C(∇∇∇′G)H · dl

(18.12.1)

H(r)= 1

jωμ

∫S

[−k2G(n× E )−((n× E )·∇∇∇′)∇∇∇′G+ jωμ(n×H )×∇∇∇′G]dS′

= − 1

jωμ∇∇∇× (∇∇∇×

∫SG(n× E )dS′

)+∇∇∇×∫SG(n×H )dS′

=∫S

[jωεG(n× E )+(n ·H )∇∇∇′G+ (n×H )×∇∇∇′G]dS′ + 1

jωμ

∮C(∇∇∇′G)E · dl

=∫S

[H∂G∂n′

−G ∂H

∂n′

]dS′ −

∮CGH× dl+ 1

jωμ

∮C(∇∇∇′G)E · dl

(18.12.2)The proof of the equivalence of these expressions is outlined in Problems 18.7 and

18.8. The Kottler-Franz formulas (18.12.1) and (18.12.2) are valid for points off theaperture surface S. The formulas are not consistent for points on the aperture. However,they have been used very successfully in practice to predict the radiation patterns ofaperture antennas.

18.13. Fresnel Diffraction 823

Fig. 18.12.1 Aperture surface S bounded by contour C.

The line-integral correction terms have a minor effect on the mainlobe and nearsidelobes of the radiation pattern. Therefore, they can be ignored and the diffractedfield can be calculated by any of the four alternative formulas, Kottler, Franz, Stratton-Chu, or Kirchhoff integral—all applied to the open surface S.

18.13 Fresnel Diffraction

In Sec. 18.4, we looked at the radiation fields arising from the Kottler-Franz formulas,where we applied the Fraunhofer approximation in which only linear phase variationsover the aperture were kept in the propagation phase factor e−jkR. Here, we considerthe intermediate case of Fresnel approximation in which both linear and quadratic phasevariations are retained.

We discuss the classical problem of diffraction of a spherical wave by a rectangularaperture, a slit, and a straight-edge using the Kirchhoff integral formula. The case of aplane wave incident on a conducting edge is discussed in Problem 18.11 using the field-equivalence principle and Kottler’s formula and more accurately, in Sec. 18.15, usingSommerfeld’s exact solution of the geometrical theory of diffraction. These examplesare meant to be an introduction to the vast subject of diffraction.

In Fig. 18.13.1, we consider a rectangular aperture illuminated from the left by a pointsource radiating a spherical wave. We take the origin to be somewhere on the apertureplane, but eventually we will take it to be the point of intersection of the aperture planeand the line between the source and observation points P1 and P2.

The diffracted field at point P2 may be calculated from the Kirchhoff formula appliedto any of the cartesian components of the field:

E =∫S

[E1∂G∂n′

−G ∂E1

∂n′

]dS′ (18.13.1)

where E1 is the spherical wave from the source point P1 evaluated at the aperture pointP′, and G is the Green’s function from P′ to P2:

E1 = A1e−jkR1

R1, G = e−jkR2

4πR2(18.13.2)


Fig. 18.13.1 Fresnel diffraction through rectangular aperture.

whereA1 is a constant. If r1 and r2 are the vectors pointing from the origin to the sourceand observation points, then we have for the distance vectors R1 and R2:

R1 = r1 − r′ , R1 = |r1 − r′| =√r2

1 − 2r1 · r′ + r′ · r′

R2 = r2 − r′ , R2 = |r2 − r′| =√r2

2 − 2r2 · r′ + r′ · r′(18.13.3)

Therefore, the gradient operator∇∇∇′ can be written as follows when it acts on a functionof R1 = |r1 − r′| or a function of R2 = |r2 − r′|:

∇∇∇′ = −R1∂∂R1

, ∇∇∇′ = −R2∂∂R2

where R1 and R2 are the unit vectors in the directions of R1 and R2. Thus, we have:

∂E1

∂n′= n ·∇∇∇′E1 = −n · R1

∂E1

∂R1= (n · R1)

(jk+ 1

R1

)A1

e−jkR1

R1

∂G∂n′

= n ·∇∇∇′G = −n · R2∂G∂R2

= (n · R2)(jk+ 1

R2

)e−jkR2

4πR2

(18.13.4)

Dropping the 1/R2 terms, we find for the integrand of Eq. (18.13.1):

E1∂G∂n′

−G ∂E1

∂n′= jkA1

4πR1R2

[(n · R2)−(n · R1)

]e−jk(R1+R2)

Except in the phase factor e−jk(R1+R2), we may replace R1 � r1 and R2 � r2, that is,

E1∂G∂n′

−G ∂E1

∂n′= jkA1

4πr1r2

[(n · r2)−(n · r1)

]e−jk(R1+R2) (18.13.5)

Thus, we have for the diffracted field at point P2:

E = jkA1

4πr1r2

[(n · r2)−(n · r1)

]∫Se−jk(R1+R2) dS′ (18.13.6)

18.13. Fresnel Diffraction 825

The quantity[(n · r2)−(n · r1)

]is an obliquity factor. Next, we set r = r1 + r2 and

define the ”free-space” field at the point P2:

E0 = A1e−jk(r1+r2)

r1 + r2= A1

e−jkr

r(18.13.7)

If the origin were the point of intersection between the aperture plane and the lineP1P2, then E0 would represent the field received at point P2 in the unobstructed casewhen the aperture and screen are absent.

The ratio D = E/E0 may be called the diffraction coefficient and depends on theaperture and the relative geometry of the points P1, P2:

D = EE0= jk

4πF[(n · r2)−(n · r1)

]∫Se−jk(R1+R2−r1−r2) dS′ (18.13.8)

where we defined the “focal length” between r1 and r2:

1

F= 1

r1+ 1

r2⇒ F = r1r2

r1 + r2(18.13.9)

The Fresnel approximation is obtained by expanding R1 and R2 in powers of r′ andkeeping only terms up to second order. We rewrite Eq. (18.13.3) in the form:

R1 = r1

√1− 2r1 · r′

r1+ r′ · r′

r21, R2 = r2

√1− 2r2 · r′

r2+ r′ · r′

r22

Next, we apply the Taylor series expansion up to second order:

√1+ x = 1+ 1

2x− 1

8x2

This gives the approximations of R1, R2, and R1 +R2 − r1 − r2:

R1 = r1 − r1 · r′ + 1

2r1

[r′ · r′ − (r1 · r′)2]

R2 = r2 − r2 · r′ + 1

2r2

[r′ · r′ − (r2 · r′)2]

R1 +R2 − r1 − r2 = −(r1 + r2)·r′ + 1

2

[(1

r1+ 1

r2

)r′ · r′ − (r1 · r′)2

r1− (r2 · r′)2

r2

]

To simplify this expression, we now assume that the origin is the point of intersectionof the line of sight P1P2 and the aperture plane. Then, the vectors r1 and r2 are anti-parallel and so are their unit vectors r1 = −r2. The linear terms cancel and the quadraticones combine to give:

R1+R2−r1−r2 = 1

2F[r′ ·r′−(r2 ·r′)2] = 1

2F∣∣r′− r2(r′ · r2)

∣∣2 = 1

2Fb′ ·b′ (18.13.10)

where we defined b′ = r′ − r2(r′ · r2), which is the perpendicular vector from the pointP′ to the line-of-sight P1P2, as shown in Fig. 18.13.1.


It follows that the Fresnel approximation of the diffraction coefficient for an arbitraryaperture will be given by:

D = EE0= jk(n · r2)

2πF

∫Se−jk(b

′·b′)/(2F) dS′ (18.13.11)

A further simplification is obtained by assuming that the aperture plane is the xy-plane and that the line P1P2 lies on the yz plane at an angle θ with the z-axis, as shownin Fig. 18.13.2.

Fig. 18.13.2 Fresnel diffraction by rectangular aperture.

Then, we have r′ = x′x + y′y, n = z, and r2 = z cosθ + y sinθ. It follows thatn · r2 = cosθ, and the perpendicular distance b′ · b′ becomes:

b′ · b′ = r′ · r′ − (r′ · r2)2= x′2 + y′2 − (y′ sinθ)2= x′2 + y′2 cos2 θ

Then, the diffraction coefficient (18.13.11) becomes:

D = jk cosθ2πF

∫ x2

−x1

∫ y2

−y1e−jk(x

′2+y′2 cos2 θ)/2F dx′dy′ (18.13.12)

where we assumed that the aperture limits are (with respect to the new origin):

−x1 ≤ x′ ≤ x2 , −y1 ≤ y′ ≤ y2

The end-points y1, y2 are shown in Fig. 18.13.2. The integrals may be expressedin terms of the Fresnel functions C(x), S(x), and F(x)= C(x)−jS(x) discussed inAppendix F. There, the complex function F(x) is defined by:

F(x)= C(x)−jS(x)=∫ x

0e−j(π/2)u

2du (Fresnel function) (18.13.13)

18.14. Knife-Edge Diffraction 827

We change integration variables to the normalized Fresnel variables:

u =√

kπF

x′ , v =√

kπF

y′ cosθ (18.13.14)

where b′ = y′ cosθ is the perpendicular distance from P′ to the line P1P2, as shown inFig. 18.13.2. The corresponding end-points are:

ui =√

kπF

xi , vi =√

kπF

yi cosθ =√

kπF

bi , i = 1,2 (18.13.15)

Note that the quantities b1 = y1 cosθ and b2 = y2 cosθ are the perpendiculardistances from the edges to the line P1P2. Since dudv = (k cosθ/πF)dx′dy′, weobtain for the diffraction coefficient:

D = j2

∫ u2

−u1

e−jπu2/2 du

∫ v2

−v1

e−jπv2/2 dv = j

2

[F(u2)−F(−u1)][F(v2)−F(−v1)

]

Noting that F(x) is an odd function and that j/2 = 1/(1− j)2, we obtain:

D = EE0= F(u1)+F(u2)

1− jF(v1)+F(v2)

1− j (rectangular aperture) (18.13.16)

The normalization factors (1−j) correspond to the infinite aperture limit u1, u2, v1,v2 → ∞, that is, no aperture at all. Indeed, since the asymptotic value of F(x) isF(∞)= (1− j)/2, we have:

F(u1)+F(u2)1− j

F(v1)+F(v2)1− j −→ F(∞)+F(∞)

1− jF(∞)+F(∞)

1− j = 1

In the case of a long slit along the x-direction, we only take the limit u1, u2 →∞:

D = EE0= F(v1)+F(v2)

1− j (diffraction by long slit) (18.13.17)

18.14 Knife-Edge Diffraction

The case of straight-edge or knife-edge diffraction is obtained by taking the limit y2 →∞, or v2 → ∞, which corresponds to keeping the lower edge of the slit. In this limitF(v2)→F(∞)= (1− j)/2. Denoting v1 by v, we have:

D(v)= 1

1− j(F(v)+1− j

2

), v =

√kπF

b1 (18.14.1)

where,

F(v)=∫ v

0e−jπu

2/2 du , D(v)= 1

1− j∫ v−∞

e−jπu2/2 du (18.14.2)


Fig. 18.14.1 Illuminated and shadow regions in straight-edge diffraction.

Positive values of v correspond to positive values of the clearance distance b1, plac-ing the point P2 in the illuminated region, as shown in Fig. 18.14.1. Negative values ofv correspond to b1 < 0, placing P2 in the geometrical shadow region behind the edge.

The magnitude-square |D|2 represents the intensity of the diffracted field relativeto the intensity of the unobstructed field. Since |1− j|2 = 2, we find:

|D(v)|2 = |E|2|E0|2 =

1

2

∣∣∣∣F(v)+1− j2

∣∣∣∣2

(18.14.3)

or, in terms of the real and imaginary parts of F(v):

|D(v)|2 = 1

2

[(C(v)+1

2

)2

+(S(v)+1

2

)2]

(18.14.4)

The quantity |D(v)|2 is plotted versus v in Fig. 18.14.2. At v = 0, corresponding tothe line P1P2 grazing the top of the edge, we haveF(0)= 0, D(0)= 1/2, and |D(0)|2 =1/4 or a 6 dB loss. The first maximum in the illuminated region occurs at v = 1.2172and has the value |D(v)|2 = 1.3704, or a gain of 1.37 dB.

The asymptotic behavior of D(v) for v → ±∞ is obtained from Eq. (F.4). We havefor large positive x:

F(±x)→ ±(

1− j2

+ jπx

e−jπx2/2)

This implies that:

D(v)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1− 1− j2πv

e−jπv2/2, for v → +∞

−1− j2πv

e−jπv2/2, for v → −∞

(18.14.5)

We may combine the two expressions into one with the help of the unit-step functionu(v) by writing D(v) in the following form, which defines the asymptotic diffractioncoefficient d(v):

D(v)= u(v)+d(v)e−jπv2/2 (18.14.6)


−3 −2 −1 0 1 2 3 4 50

0.25

0.5

0.75

1

1.25

1.5

|D

(ν)|

2

ν

Diffraction Coefficient

−3 −2 −1 0 1 2 3 4 5−24

−18

−12

−6

0

20lo

g 10|

D(ν

)|

ν

Diffraction Coefficient in dB

Fig. 18.14.2 Diffraction coefficient in absolute and dB units.

where u(v)= 1 for v ≥ 0 and u(v)= 0 for v < 0.With u(0)= 1, this definition requires d(0)= D(0)−v(0)= 0.5 − 1 = −0.5. But if

we define u(0)= 0.5, as is sometimes done, then, d(0)= 0. The asymptotic behavior ofD(v) can now be expressed in terms of the asymptotic behavior of d(v):

d(v)= −1− j2πv

, for v → ±∞ (18.14.7)

In the illuminated region D(v) tends to unity, whereas in the shadow region it de-creases to zero with asymptotic dB attenuation or loss:

L = −10 log10

∣∣d(v)∣∣2 = 10 log10

(2π2v2) , as v → −∞ (18.14.8)

The MATLAB function diffr.m, mentioned in Appendix F, calculates the diffractioncoefficient (18.14.1) at any vector of (real) values of v. It has usage:

D = diffr(v); % knife-edge diffraction coefficient D(v)

For values v ≤ 0.7, the diffraction loss can be approximated very well by the follow-ing function [1309]:

L = −10 log10

∣∣D(v)∣∣2 = 6.9+ 20 log10

(√(v+ 0.1)2+1− v− 0.1

)(18.14.9)

Example 18.14.1: Diffraction Loss over Obstacles. The propagation path loss over obstacles andirregular terrain is usually determined using knife-edge diffraction. Fig. 18.14.3 illustratesthe case of two antennas communicating over an obstacle. For small angles θ, the focallength F is often approximated in several forms:

F = r1r2

r1 + r2� d1d2

d1 + d2� l1l2l1 + l2

These approximations are valid typically when d1, d2 are much greater than λ and theheight h of the obstacle, typically, at least ten times greater. The clearance distance can


Fig. 18.14.3 Communicating antennas over an obstacle.

be expressed in terms of the heights:

b1 = y1 cosθ =(h1d2 + h2d1

d1 + d2− h

)cosθ

The distance b1 can also be expressed approximately in terms of the subtended anglesα1,α2, and α, shown in Fig. 18.14.3:

b1 � l1α1 � l2α2 ⇒ b1 =√l1l2α1α2 (18.14.10)

and in terms of α, we have:

α1 = αl2l1 + l2 , α2 = αl1

l1 + l2 ⇒ b1 = αF ⇒ v = α√

2Fλ

(18.14.11)

The case of multiple obstacles has been studied using appropriate modifications of theknife-edge diffraction problem and the geometrical theory of diffraction [1398–1413]. ��

Example 18.14.2: Fresnel Zones. Consider two antennas separated by a distance d and an ob-stacle at distance z from the midpoint with clearance b, as shown below. Fresnel zones andthe corresponding Fresnel zone ellipsoids help answer the question of what the minimumvalue of the clearance b should be for efficient communication between the antennas.

0 1 2 3 4 5−3

−2

−1

0

1

2

3

20lo

g 10|

D(ν

)|

ν

Diffraction Coefficient in dB

exact asymptotic extrema fresnel zone


The diffraction coefficient D(v) and its asymptotic form were given in Eqs. (18.14.1) and(18.14.5), that is,

D(v)= 1

1− j(F(v)+1− j

2

), v =

√kπF

b =√

2

λFb , F = d1d2

d1 + d2(18.14.12)

and for positive and large clearance b, or equivalently, for large positive v,

Das(v)= 1− 1− j2πv

e−jπv2/2 = 1− 1√

2πve−jπ(v

2/2+1/4) (18.14.13)

As can be seen in the above figure on the right, the diffraction coefficients D(v) andDas(v) agree closely even for small values of v. Therefore, the extrema can be obtainedfrom the asymptotic form. They correspond to the values of v that cause the exponentialin (18.14.13) to take on its extremal values of±1, that is, the v’s that satisfy v2/2+1/4 = n,with integer n, or:

vn =√

2n− 0.5 , n = 1,2, . . . (18.14.14)

The corresponding values of D(v), shown on the figure with black dots, are given by

Das(vn)= 1− 1√2πvn

e−jπn = 1− 1√2πvn

(−1)n (18.14.15)

An alternative set of v’s, also corresponding to alternating almost extremum values, arethose that define the conventional Fresnel zones, that is,

un =√

2n , n = 1,2, . . . (18.14.16)

These are indicated by open circles on the graph. The corresponding D(v) values are:

Das(un)= 1− e−jπ/4√2πun

(−1)n (18.14.17)

For clearances b that correspond to v’s that are too small, i.e., v < 0.5, the diffractioncoefficient D(v) becomes too small, impeding efficient communication. The smallest ac-ceptable clearance b is taken to correspond to the first maximum of D(v), that is, v = v1

or more simply v = u1 =√

2.

The locus of points (b, z) corresponding to a fixed value of v, and hence to a fixed valueof the diffraction coefficient D(v), form an ellipsoid. This can be derived from (18.14.12)by setting d1 = d/2+ z and d2 = d/2− z, that is,

v =√

2

λFb ⇒ b2 = λF

2v2 = λ(d2/4− z2)

2dv2 , because F = d1d2

d1 + d2= d2/4− z2

d

which can be rearranged into the equation of an ellipse:(8

v2λd

)b2 +

(4

d2

)z2 = 1

For v = u1 =√

2, this defines the first Fresnel zone ellipse, which gives the minimumacceptable clearance for a given distance z:(

4

λd

)b2 +

(4

d2

)z2 = 1 (18.14.18)


If the obstacle is at midpoint (z = 0), the minimum clearance becomes:

b = 1

2

√λd (18.14.19)

For example, for a distance of d = 1 km, using a cell phone frequency of f = 1 GHz,corresponding to wavelength λ = 30 cm, we find b = √λd/2 = 8.66 meters.

A common interpretation and derivation of Fresnel zones is to consider the path differencebetween the rays following the straight path connecting the two antennas and the pathgetting scattered from the obstacle, that is, Δl = l1 + l2 − d. From the indicated triangles,and assuming that b� d1 and b� d2, we find:

l1 =√d2

1 + b2 � d1 + b2

2d1, l2 =

√d2

2 + b2 � d2 + b2

2d2

which leads to the following path length Δl, expressed in terms of v:

Δl = l1 + l2 − d = b2

2

(1

d1+ 1

d2

)= b2

2F= λ

4v2

The corresponding phase difference between the two paths, e−jkΔl, will be then:

e−jkΔl = e−jπv2/2 (18.14.20)

which has the same form as in the diffraction coefficient Das(v). The values v = un =√2n will make the path difference a multiple of λ/2, that is, Δl = nλ/2, resulting in the

alternating phase e−jkΔl = (−1)n.

The discrepancy between the choices vn and un arises from using D(v) to find the alter-nating maxima, versus using the plain phase (18.14.20). ��

The Fresnel approximation is not invariant under shifting the origin. Our choice oforigin above is not convenient because it depends on the observation point P2. If wechoose a fixed origin, such as the point O in Fig. 18.14.4, then, we must determine thecorresponding Fresnel coefficient.

We assume that the points P1, P2 lie on the yz plane and take P2 to lie in the shadowregion. The angles θ1, θ2 may be chosen to be positive or negative to obtain all possiblelocations of P1, P2 relative to the screen.

The diffraction coefficient is still given by Eq. (18.13.8) but with r1, r2 replaced bythe distances l1, l2. The unit vectors towards P1 and P2 are:

l1 = −z cosθ1 − y sinθ1 , l2 = z cosθ2 − y sinθ2 (18.14.21)

Since r′ = x′x+ y′y and n = z, we find:

l1 · r′ = −y′ sinθ1 , l2 · r′ = −y′ sinθ2 , n · l1 = − cosθ1 , n · l2 = cosθ2

The quadratic approximation for the lengths R1, R2 gives, then:


Fig. 18.14.4 Fresnel diffraction by straight edge.

R1 +R2 − l1 − l2 = −(l1 + l2)·r′ + 1

2

[(1

l1+ 1

l2

)(r′ · r′)− (l1 · r′)2

l1− (l2 · r′)2

l2

]

= y′(sinθ1 + sinθ2)+(

1

l1+ 1

l2

)x′2

2+(

cos2 θ1

l1+ cos2 θ2

l2

)y′2

2

= 1

2Fx′2 + 1

2F′[y′2 + 2F′y′(sinθ1 + sinθ2)

]

= 1

2Fx′2 + 1

2F′(y′ + y0)2− 1

2F′y2

0

where we defined the focal lengths F,F′ and the shift y0:

1

F= 1

l1+ 1

l2,

1

F′= cos2 θ1

l1+ cos2 θ2

l2, y0 = F′(sinθ1 + sinθ2) (18.14.22)

Using these approximations in Eq. (18.13.6) and replacing r1, r2 by l1, l2, we find:

E = jkA1e−jk(l1+l2)

4πl1l2

[(n · l2)−(n · l1)

]∫Se−jk(R1+R2−l1−l2) dS′

= jkA1e−k(l1+l2)

4πl1l2(cosθ1 + cosθ2)ejky

20/2F′

∫e−jkx

′2/2F−jk(y′+y0)2/2F′ dx′dy′

The x′-integral is over the range −∞ < x′ < ∞ and can be converted to a Fresnelintegral with the change of variables u = x′√k/(πF):∫∞

−∞e−jkx

′2/2F dx′ =√πFk

∫∞−∞e−jπu

2/2 du =√πFk(1− j)

The y′-integral is over the upper-half of the xy-plane, that is, 0 ≤ y′ < ∞. Definingthe Fresnel variables u = (y′ + y0)

√k/(πF′) and v = y0

√k/(πF′), we find:

∫∞0e−jk(y

′+y0)2/2F′ dy′ =√πF′

k

∫∞ve−jπu

2/2 du =√πF′

k(1− j)D(−v)


where the function D(v) was defined in Eq. (18.14.1). Putting all the factors together,we may write the diffracted field at the point P2 in the form:

E = Eedgee−jkl2√l2Dedge (straight-edge diffraction) (18.14.23)

where we set ky20/2F′ = πv2/2 and defined the incident field Eedge at the edge and the

overall edge-diffraction coefficient Dedge by:

Eedge = A1e−jkl1l1

, Dedge =√FF′

l2

(cosθ1 + cosθ2

2

)ejπv

2/2D(−v) (18.14.24)

The second factor (e−jkl2/√l2) in (18.14.23) may be interpreted as a cylindrical wave

emanating from the edge as a result of the incident field Eedge. The third factor Dedge isthe angular gain of the cylindrical wave. The quantity v may be written as:

v =√

kπF′

y0 =√kF′

π(sinθ1 + sinθ2) (18.14.25)

Depending on the sign and relative sizes of the angles θ1 and θ2, it follows thatv > 0 when P2 lies in the shadow region, and v < 0 when it lies in the illuminatedregion. For large positive v, we may use Eq. (18.14.5) to obtain the asymptotic form ofthe edge-diffraction coefficient Dedge:

Dedge =√FF′

l2cosθ1 + cosθ2

2ejπv

2/2 1− j2πv

e−jπv2/2 =

√FF′

l2cosθ1 + cosθ2

2

1− j2πv

Writing√F/l2 =

√l1/(l1 + l2) and replacing v from Eq. (18.14.25), the

√F′ factor

cancels and we obtain:

Dedge =√

l1l1 + l2

(1− j)(cosθ1 + cosθ2)4√πk(sinθ1 + sinθ2)

(18.14.26)

This expression may be simplified further by defining the overall diffraction angleθ = θ1 + θ2, as shown in Fig. 18.14.4 and using the trigonometric identity:

cosθ1 + cosθ2

sinθ1 + sinθ2= cot

(θ1 + θ2

2

)

Then, Eq. (18.14.26) may be written in the form:

Dedge =√

l1l1 + l2

(1− j)4√πk

cotθ2

(18.14.27)

The asymptotic diffraction coefficient is obtained from Eqs. (18.14.26) or (18.14.27)by taking the limit l1 →∞, which gives

√l1/(l1 + l2)→ 1. Thus,

Dedge = (1− j)(cosθ1 + cosθ2)4√πk(sinθ1 + sinθ2)

= (1− j)4√πk

cotθ2

(18.14.28)

18.15. Geometrical Theory of Diffraction 835

Eqs. (18.14.27) and (18.14.28) are equivalent to those given in [1302].The two choices for the origin lead to two different expressions for the diffracted

fields. However, the expressions agree near the forward direction, θ � 0. It is easilyverified that both Eq. (18.14.1) and (18.14.27) lead to the same approximation for thediffracted field:

E = Eedgee−jkl2√l2

√l1

l1 + l21− j

2√πkθ

(18.14.29)

18.15 Geometrical Theory of Diffraction

Geometrical theory of diffraction is an extension of geometrical optics [1398–1413]. Itviews diffraction as a local edge effect. In addition to the ordinary rays of geometricaloptics, it postulates the existence of “diffracted rays” from edges. The diffracted rayscan reach into shadow regions, where geometrical optics fails.

An incident ray at an edge generates an infinity of diffracted rays emanating from theedge having different angular gains given by a diffraction coefficientDedge. An exampleof such a diffracted ray is given by Eq. (18.14.23).

The edge-diffraction coefficient Dedge depends on (a) the type of the incident wave,such as plane wave, or spherical, (b) the type and local geometry of the edge, such as aknife-edge or a wedge, and (c) the directions of the incident and diffracted rays.

The diffracted field and coefficient are usually taken to be in their asymptotic forms,like those of Eq. (18.15.25). The asymptotic forms are derived from certain exactlysolvable canonical problems, such as a conducting edge, a wedge, and so on.

The first and most influential of all such problems was Sommerfeld’s solution of aplane wave incident on a conducting half-plane [1287], and we discuss it below.

Fig. 18.15.1 shows a plane wave incident at an angle α on the conducting planeoccupying half of the xz-plane for x ≥ 0. The plane of incidence is taken to be the xy-plane. Because of the cylindrical symmetry of the problem, we may assume that thereis no z-dependence and that the fields depend only on the cylindrical coordinates ρ,φ.

Fig. 18.15.1 Plane wave incident on conducting half-plane.


Two polarizations may be considered: TE, in which the electric field is E = zEz, andTM, which has H = zHz. Using cylindrical coordinates defined in Eq. (E.2) of AppendixE, and setting ∂/∂z = 0, Maxwell’s equations reduce in the two cases into:

(TE) ∇2Ez + k2Ez = 0, Hρ = − 1

jωμ1

ρ∂Ez∂φ

, Hφ = 1

jωμ∂Ez∂ρ

(TM) ∇2Hz + k2Hz = 0, Eρ = 1

jωε1

ρ∂Hz∂φ

, Eφ = − 1

jωε∂Hz∂ρ

(18.15.1)

where k2 =ω2με, and the two-dimensional∇∇∇2 is in cylindrical coordinates:

∇2 = 1

ρ∂∂ρ

(ρ∂∂ρ

)+ 1

ρ2

∂2

∂φ2(18.15.2)

The boundary conditions require that the tangential electric field be zero on bothsides of the conducting plane, that is, for φ = 0 and φ = 2π. In the TE case, thetangential electric field is Ez, and in the TM case, Ex = Eρ cosφ − Eφ sinφ = Eρ =(1/jωερ)(∂Hz/∂φ), for φ = 0,2π. Thus, the boundary conditions are:

(TE) Ez = 0, for φ = 0 and φ = 2π

(TM)∂Hz∂φ

= 0, for φ = 0 and φ = 2π(18.15.3)

In Fig. 18.15.1, we assume that 0 ≤ α ≤ 90o and distinguish three wedge regionsdefined by the half-plane and the directions along the reflected and transmitted rays:

reflection region (AOB): 0 ≤ φ ≤ π−αtransmission region (BOC): π−α ≤ φ ≤ π+αshadow region (COA): π+α ≤ φ ≤ 2π

(18.15.4)

The case when 90o ≤ α ≤ 180o is shown in Fig. 18.15.2, in which α has beenredefined to still be in the range 0 ≤ α ≤ 90o. The three wedge regions are now:

reflection region (AOB): 0 ≤ φ ≤ αtransmission region (BOC): α ≤ φ ≤ 2π−αshadow region (COA): 2π−α ≤ φ ≤ 2π

(18.15.5)

We construct the Sommerfeld solution in stages. We start by looking for solutionsof the Helmholtz equation∇2U+k2U = 0 that have the factored form: U = ED, whereE is also a solution, but a simple one, such as that of the incident plane wave. Using thedifferential identities of Appendix C, we have:

∇2U + k2U = D(∇2E + k2E)+ E∇2D+ 2∇∇∇E ·∇∇∇D

Thus, the conditions ∇2U + k2U = 0 and ∇2E + k2E = 0 require:

E∇2D+ 2∇∇∇E ·∇∇∇D = 0 ⇒ ∇2D+ 2(∇∇∇ lnE)·∇∇∇D = 0 (18.15.6)


Fig. 18.15.2 Plane wave incident on conducting half-plane.

If we assume that E is of the form E = ejf , where f is a real-valued function, then,equating to zero the real and imaginary parts of ∇2E + k2E = 0, we find for f :

∇2E + k2E = E(k2 −∇∇∇f ·∇∇∇f + j∇2f) = 0 ⇒ ∇2f = 0 , ∇∇∇f ·∇∇∇f = k2 (18.15.7)

Next, we assume that D is of the form:

D = D0

∫ v−∞e−jg(u)du (18.15.8)

where D0 is a constant, v is a function of ρ,φ, and g(u) is a real-valued function to bedetermined. Noting that ∇∇∇D = D0e−jg∇∇∇v and ∇∇∇g = g′(v)∇∇∇v, we find:

∇∇∇D = D0e−jg∇∇∇v , ∇2D = D0e−jg(∇2v− jg′(v)∇∇∇v ·∇∇∇v)

Then, it follows from Eq. (18.15.6) that ∇2D+ 2(∇∇∇ lnE)·∇∇∇D = ∇2D+ j∇∇∇f ·∇∇∇D and:

∇2D+ j∇∇∇f ·∇∇∇D = D0e−jg[∇2v+ j(2∇∇∇f ·∇∇∇v− g′∇∇∇v ·∇∇∇v)]= 0

Equating the real and imaginary parts to zero, we obtain the two conditions:

∇2v = 0 ,2∇∇∇f ·∇∇∇v∇∇∇v ·∇∇∇v = g′(v) (18.15.9)

Sommerfeld’s solution involves the Fresnel diffraction coefficient of Eq. (18.14.1),which can be written as follows:

D(v)= 1

1− j[

1− j2

+F(v)]= 1

1− j∫ v−∞e−jπu

2/2du (18.15.10)

Therefore, we are led to choose g(u)= πu2/2 and D0 = 1/(1− j). To summarize,we may construct a solution of the Helmholtz equation in the form:

∇2U + k2U = 0 , U = ED = ejfD(v) (18.15.11)


where f and v must be chosen to satisfy the four conditions:

∇2f = 0, ∇∇∇f ·∇∇∇f = k2

∇2v = 0,2∇∇∇f ·∇∇∇v∇∇∇v ·∇∇∇v = g′(v)= πv

(18.15.12)

It can be verified easily that the functions u = ρa cosaφ and u = ρa sinaφ are solu-tions of the two-dimensional Laplace equation∇2u = 0, for any value of the parametera. Taking f to be of the form f = Aρa cosaφ, we have the condition:

∇∇∇f = Aaρa−1[ρρρ cosaφ− φφφ sinaφ] ⇒ ∇∇∇f ·∇∇∇f = A2a2ρ2(a−1) = k2

This immediately implies that a = 1 and A2 = k2, so that A = ±k. Thus, f =Aρ cosφ = ±kρ cosφ. Next, we choose v = Bρa cosaφ. Then:

∇∇∇f = A(ρρρ cosφ− φφφ sinφ)

∇∇∇v = Baρa−1[ρρρ cosaφ− φφφ sinaφ]

∇∇∇f ·∇∇∇v = ABaρa−1[cosφ cosaφ+ sinφ sinaφ] = ABaρa−1 cos(φ− aφ)

∇∇∇v ·∇∇∇v = B2a2ρ2(a−1)

Then, the last of the conditions (18.15.12) requires that:

1

πv2∇∇∇f ·∇∇∇v∇∇∇v ·∇∇∇v = 2Aρ1−2a cos(φ− aφ)

πaB2 cosaφ= 1

which implies that a = 1/2 and B2 = 2A/πa = 4A/π. But since A = ±k, only thecase A = k is compatible with a real coefficient B. Thus, we have B2 = 4k/π, or,B = ±2

√k/π.

In a similar fashion, we find that if we take v = Bρa sinaφ, then a = 1/2, but nowB2 = −4A/π, requiring that A = −k, and B = ±2

√k/π. In summary, we have the

following solutions of the conditions (18.15.12):

f = +kρ cosφ, v = ±2

√kπρ1/2 cos

φ2

f = −kρ cosφ, v = ±2

√kπρ1/2 sin

φ2

(18.15.13)

The corresponding solutions (18.15.11) of the Helmholtz equation are:

U(ρ,φ)= ejkρ cosφD(v) , v = ±2

√kπρ1/2 cos

φ2

U(ρ,φ)= e−jkρ cosφD(v) , v = ±2

√kπρ1/2 sin

φ2

(18.15.14)


The function D(v) may be replaced by the equivalent form of Eq. (18.14.6) in orderto bring out its asymptotic behavior for large v:

U(ρ,φ)= ejkρ cosφ[u(v)+d(v)e−jπv2/2], v = ±2

√kπρ1/2 cos

φ2

U(ρ,φ)= e−jkρ cosφ[u(v)+d(v)e−jπv2/2], v = ±2

√kπρ1/2 sin

φ2

Using the trigonometric identities cosφ = 2 cos2(φ/2)−1 = 1 − 2 sin2(φ/2), wefind for the two choices of v:

kρ cosφ− 1

2πv2 = kρ

[cosφ− 2 cos2 φ

2

]= −kρ

−kρ cosφ− 1

2πv2 = −kρ

[cosφ+ 2 sin2 φ

2

]= −kρ

Thus, an alternative form of Eq. (18.15.14) is:

U(ρ,φ)= ejkρ cosφ u(v)+e−jkρ d(v) , v = ±2

√kπρ1/2 cos

φ2

U(ρ,φ)= e−jkρ cosφ u(v)+e−jkρ d(v) , v = ±2

√kπρ1/2 sin

φ2

(18.15.15)

Shifting the origin of the angle φ still leads to a solution. Indeed, defining φ′ =φ±α, we note the property ∂/∂φ′ = ∂/∂φ, which implies the invariance of the Laplaceoperator under this change. The functions U(ρ,φ ± α) are the elementary solutionsfrom which the Sommerfeld solution is built.

Considering the TE case first, the incident plane wave in Fig. 18.15.1 is E = zEi,where Ei = E0e−jk·r, with r = xρ cosφ + yρ sinφ and k = −k(x cosα + y sinα). Itfollows that:

k · r = −kρ(cosφ cosα+ sinφ sinα)= −kρ cos(φ−α)Ei = E0e−jk·r = E0ejkρ cos(φ−α) (18.15.16)

The image of this electric field with respect to the perfect conducting plane willbe the reflected field Er = −E0e−jkr·r, where kr = k(−x cosα + y sinα), resulting inEr = −E0ejkρ cos(φ+α). The sum Ei + Er does vanish for φ = 0 and φ = 2π, but it alsovanishes for φ = π. Therefore, it is an appropriate solution for a full conducting plane(the entire xz-plane), not for the half-plane.

Sommerfeld’s solution, which satisfies the correct boundary conditions, is obtainedby forming the linear combinations of the solutions of the type of Eq. (18.15.14):

Ez = E0[ejkρ cosφi D(vi)−ejkρ cosφr D(vr)

](TE) (18.15.17)

where

φi = φ−α, vi = 2

√kπρ1/2 cos

φi2

φr = φ+α, vr = 2

√kπρ1/2 cos

φr2

(18.15.18)


For the TM case, we form the sum instead of the difference:

Hz = H0[ejkρ cosφi D(vi)+ejkρ cosφr D(vr)

](TM) (18.15.19)

The boundary conditions (18.15.3) are satisfied by both the TE and TM solutions.As we see below, the choice of the positive sign in the definitions of vi and vr wasrequired in order to produce the proper diffracted field in the shadow region. Using thealternative forms (18.15.15), we separate the terms of the solution as follows:

Ez = E0ejkρ cosφi u(vi)−E0ejkρ cosφr u(vr)+E0e−jkρ[d(vi)−d(vr)

](18.15.20)

The first two terms correspond to the incident and reflected fields. The third term isthe diffracted field. The algebraic signs of vi and vr are as follows within the reflection,transmission, and shadow regions of Eq. (18.15.4):

reflection region: 0 ≤ φ < π−α, vi > 0, vr > 0transmission region: π−α < φ < π+α, vi > 0, vr < 0shadow region: π+α < φ ≤ 2π, vi < 0, vr < 0

(18.15.21)

The unit-step functions will be accordingly present or absent resulting in the follow-ing fields in these three regions:

reflection region: Ez = Ei + Er + Edtransmission region: Ez = Ei + Edshadow region: Ez = Ed

(18.15.22)

where we defined the incident, reflected, and diffracted fields:

Ei = E0ejkρ cosφi

Er = −E0ejkρ cosφr

Ed = E0e−jkρ[d(vi)−d(vr)

](18.15.23)

The diffracted field is present in all three regions, and in particular it is the only onein the shadow region. For large vi and vr (positive or negative), we may replace d(v) byits asymptotic form d(v)= −(1− j)/(2πv) of Eq. (18.14.7), resulting in the asymptoticdiffracted field:

Ed = −E0e−jkρ1− j2π

(1

vi− 1

vr

)

= −E0e−jkρ1− j

2π2√k/πρ1/2

(1

cos(φi/2)− 1

cos(φr/2)

)

which can be written in the form:

Ed = E0e−jkρ

ρ1/2 Dedge (18.15.24)

18.16. Problems 841

with an edge-diffraction coefficient:

Dedge = − 1− j4√πk

⎛⎜⎜⎝ 1

cosφi2

− 1

cosφr2

⎞⎟⎟⎠ (18.15.25)

Using a trigonometric identity, we may write Dedge as follows:

Dedge = − 1− j4√πk

⎛⎜⎜⎝ 1

cosφ−α

2

− 1

cosφ+α

2

⎞⎟⎟⎠ = −1− j√

πk

sinφ2

sinα2

cosφ+ cosα(18.15.26)

Eqs. (18.15.22) and (18.15.24) capture the essence of the geometrical theory of diffrac-tion: In addition to the ordinary incident and reflected geometric optics rays, one alsohas diffracted rays in all directions corresponding to a cylindrical wave emanating fromthe edge with a directional gain of Dedge.

For the case of Fig. 18.15.2, the incident and reflected plane waves have propagationvectors k = k(z cosα − y sinα) and kr = k(z cosα + y sinα). These correspond tothe incident and reflected fields:

Ei = E0e−jk·r = E0e−jkρ cos(φ+α) , Er = −E0e−jkr·r = −E0e−jkρ cos(φ−α)

In this case, the Sommerfeld TE and TM solutions take the form:

Ez = E0[e−jkρ cosφi D(vi)−e−jkρ cosφr D(vr)

]Hz = H0

[e−jkρ cosφi D(vi)+e−jkρ cosφr D(vr)

] (18.15.27)

where, now:

φi = φ+α, vi = 2

√kπρ1/2 sin

φi2

φr = φ−α, vr = −2

√kπρ1/2 sin

φr2

(18.15.28)

The choice of signs in vi and vr are such that they are both negative within theshadow region defined by Eq. (18.15.5). The same solution can also be obtained fromFig. 18.15.1 and Eq. (18.15.17) by replacing α by π−α.

18.16 Problems

18.1 Show that Eq. (18.4.9) can be written in the compact vectorial form:

E = −jk e−jkr

4πrr× [z× f− η r× (z× g)

], H = − jk

ηe−jkr

4πrr× [r× (z× f)+η z× g

]Similarly, show that Eqs. (18.4.10) and (18.4.11) can be written as:

E = −2jke−jkr

4πrr× [z× f

], H = −2jk

ηe−jkr

4πrr× [r× (z× f)

]

E = 2jkηe−jkr

4πrr× [r× (z× g)

], H = −2jk

e−jkr

4πrr× [z× g

]


18.2 Prove the first pair of equations for E,H of the previous problem by working exclusivelywith the Kottler formulas (18.4.2) and taking their far-field limits.

18.3 Explain in detail how the inequality (18.6.12) for the aperture efficiency ea may be thoughtof as an example of the Schwarz inequality. Then, using standard properties of Schwarzinequalities, prove that the maximum of ea is unity and is achieved for uniform apertures.As a reminder, the Schwarz inequality for single-variable complex-valued functions is:

∣∣∣∣∣∫ baf∗(x)g(x)dx

∣∣∣∣∣2

≤∫ ba|f(x)|2 dx ·

∫ ba|g(x)|2 dx

18.4 To prove the equivalence of the Kirchhoff diffraction and Stratton-Chu formulas, (18.10.6)and (18.10.7), use the identities (C.29) and (C.32) of Appendix C, to obtain:∫

V

[jωμG J+ 1

εG∇∇∇′ρ+G∇∇∇′ × Jm

]dV′ =

∫V

[jωμG J− ρ

ε∇∇∇′G+ Jm ×∇∇∇′G

]dV′

−∮S

[nρεG+ n× JmG

]dS′

Then, using the identity (C.33), show that Eq. (18.10.6) can be rewritten in the form:

E(r)= −∫V

[jωμG J− ρ

ε∇∇∇′G+ Jm ×∇∇∇′G

]dV′

+∮S

[nρεG+ n× JmG

]dS′

−∮S

[nG∇∇∇′ · E− (n× E)×∇∇∇′G−G n× (∇∇∇′ × E)−(n · E)∇∇∇′G]dS′

Finally, use ρ/ε =∇∇∇′ · E and ∇∇∇′ × E+ Jm = −jωμH to obtain (18.10.7).

18.5 Prove the equivalence of the Stratton-Chu and Kottler formulas, (18.10.7) and (18.10.10), byfirst proving and then using the following dual relationships:

∫V

[jωρ∇∇∇′G− (J ·∇∇∇′)∇∇∇′G]dV′ =

∮S

[((n×H)·∇∇∇′)∇∇∇′G− jωε(n · E)∇∇∇′G]

∫V

[jωρm∇∇∇′G− (Jm ·∇∇∇′)∇∇∇′G

]dV′ = −

∮S

[((n× E)·∇∇∇′)∇∇∇′G+ jωμ(n ·H)∇∇∇′G]

To prove these, work component-wise, use Maxwell’s equations (18.2.1), and apply the di-vergence theorem on the volume V of Fig. 18.10.1.

18.6 Prove the equivalence of the Kottler and Franz formulas, (18.10.10) and (18.10.11), by usingthe identity ∇∇∇ × (∇∇∇ × A)= ∇∇∇(∇∇∇ · A)−∇2A, and by replacing the quantity k2G(r − r′) by−δ(3)(r−r′)−∇′2G. Argue that the term δ(3)(r−r′)makes a difference only for the volumeintegrals, but not for the surface integrals.

18.7 Prove the equivalence of the modified Stratton-Chu and Kirchhoff diffraction integral for-mulas of Eq. (18.12.1) and (18.12.2) by using the identity (C.42) of Appendix C and replacing∇∇∇′ · E = 0 and∇∇∇′ × E = −jωμH in the source-less region under consideration.

18.8 Prove the equivalence of the Kottler and modified Stratton-Chu formulas of Eq. (18.12.1) and(18.12.2) by subtracting the two expressions, replacing jωεE =∇∇∇′×H , and using the Stokesidentity (C.38) of Appendix C.

18.16. Problems 843

18.9 Consider a reflector antenna fed by a horn, as shownon the right. A closed surface S = Sr + Sa is suchthat the portion Sr caps the reflector and the portionSa is an aperture in front of the reflector. The feedlies outside the closed surface, so that the volume Venclosed by S is free of current sources.

Applying the Kottler version of the extinction theorem of Sec. 18.10 on the volume V, showthat for points r outsideV, the field radiated by the induced surface currents on the reflectorSr is equal to the field radiated by the aperture fields on Sa, that is,

E rad(r) = 1

jωε

∫Sr

[k2G J s +

(J s ·∇∇∇′

)∇∇∇′G]dS′

= 1

jωε

∫Sa


where the induced surface currents on the reflector are J s = nr ×H and Jms = −nr ×E, andon the perfectly conducting reflector surface, we must have Jms = 0.

This result establishes the equivalence of the so-called aperture-field and current-distributionmethods for reflector antennas [1686]. See also Sec. 21.9.

18.10 Consider an x-polarized uniform plane wave incident obliquely on the straight-edge apertureof Fig. 18.14.4, with a wave vector direction k1 = z cosθ1 + y sinθ1. First show that thetangential fields at an aperture point r′ = x′ x+y′ y on the aperture above the straight-edgeare given by:

Ea = xE0e−jky′ sinθ1 , Ha = y

E0

η0cosθ1e−jky

′ sinθ1

Then, using Kottler’s formula (18.12.1), and applying the usual Fresnel approximations inthe integrand, as was done for the point source in Fig. 18.14.4, show that the diffractedwave below the edge is given by Eqs. (18.14.23)–(18.14.25), except that the field at the edgeis Eedge = E0, and the focal lengths are in this case F = l2 and F′ = l2/ cos2 θ2

Finally, show that the asymptotic diffracted field (when l2 → ∞), is given near the forwarddirection θ � 0 by:

E = Eedgee−jkl2√l2

1− j2√πkθ

18.11 Assume that the edge in the previous problem is a perfectly conducting screen. Using thefield-equivalence principle with effective current densities on the aperture above the edgeJ s = 0 and Jms = −2n× Ea, and applying the usual Fresnel approximations, show that thediffracted field calculated by Eq. (18.4.1) is still given by Eqs. (18.14.23)–(18.14.25), exceptthat the factor cosθ1+ cosθ2 is replaced now by 2 cosθ2, and that the asymptotic field andedge-diffraction coefficient are:

E = E0e−jkl2√l2Dedge , Dedge = (1− j)2 cosθ2

4√πk(sinθ1 + sinθ2)

Show that this expression agrees with the exact Sommerfeld solution (18.15.26) at normalincidence and near the forward diffracted direction.

19Diffraction – Plane-Wave Spectrum

This chapter continues the previous one on radiation from apertures. The emphasisis on Rayleigh-Sommerfeld diffraction theory, plane-wave spectrum representation forscalar and vector fields, radiated and reactive power of apertures, integral equationsfor apertures in conducting screens, revisiting the Sommerfeld half-plane problem us-ing Wiener-Hopf factorization techniques, the Bethe-Bouwkamp model of diffraction bysmall holes, and the Babinet principle.

19.1 Rayleigh-Sommerfeld Diffraction Theory

In this section, we recast Kirchhoff’s diffraction formula in a form that uses a Dirich-let Green’s function (i.e., one that vanishes on the boundary surface) and obtain theRayleigh-Sommerfeld diffraction formula. In the next section, we show that this re-formulation is equivalent to the plane-wave spectrum approach to diffraction, and useit to justify the modified forms (18.1.2) and (18.1.3) of the field equivalence principle.In Chap. 20, we use it to obtain the usual Fresnel and Fraunhofer approximations anddiscuss a few applications from Fourier optics.

We will work with the scalar case, but the same method can be used for the vectorcase. With reference to Fig. 19.1.1, we consider a scalar field E(r) that satisfies thesource-free Helmholtz equation, (∇2 + k2)E(r)= 0, over the right half-space z ≥ 0.

We consider a closed surface consisting of the surface S∞ of a sphere of very largeradius centered at the observation point r and bounded on the left by its intersection Swith the xy plane, as shown in the Fig. 19.1.1. Clearly, in the limit of infinite radius, thevolume V bounded by S+S∞ is the right half-space z ≥ 0, and S becomes the entire xyplane. Applying Eq. (18.10.3) to volume V, we have:

∫V

[G(∇′2E + k2E)−E (∇′2G+ k2G)

]dV′ = −

∮S+S∞

[G∂E∂n′

− E ∂G∂n′

]dS′ (19.1.1)

The surface integral over S∞ can be ignored by noting that n is the negative of theradial unit vector and therefore, we have after adding and subtracting the term jkEG:

−∫S∞

[G∂E∂n′

− E ∂G∂n′

]dS′ =

∫S∞

[G(∂E∂r

+ jkE)− E

(∂G∂r

+ jkG)]dS′

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800 18. Radiation from Apertures - Rutgers ECEorfanidi/ewa/ch18.pdf · 18 Radiation from Apertures 18.1 Field Equivalence Principle The radiation ﬁelds from aperture antennas, such